What is the total combination for choosing 8 questions out of 10?

  • Context: High School 
  • Thread starter Thread starter caesarkim
  • Start date Start date
  • Tags Tags
    Probability
Click For Summary
SUMMARY

The total number of combinations for choosing 8 questions out of 10 is 45, calculated using the formula 10! / (8! * 2!). This accounts for the fact that there are 10 choices for the first question to leave out and 9 for the second, resulting in 90 combinations. Since the order of selection does not matter, the result is divided by 2 to correct for double counting. This is a fundamental problem in combinatorics, represented as _nC_r.

PREREQUISITES
  • Understanding of factorial notation and operations
  • Basic knowledge of combinatorics
  • Familiarity with mathematical notation for combinations
  • Ability to use scientific calculators or software for calculations
NEXT STEPS
  • Learn about permutations and their differences from combinations
  • Explore advanced combinatorial problems and their applications
  • Study the use of combinations in probability theory
  • Practice using calculators or programming languages to compute combinations
USEFUL FOR

Students, educators, and professionals in mathematics, statistics, and fields requiring combinatorial analysis will benefit from this discussion.

caesarkim
there are 10 questions. i can choose only exact 8 questions.

what is the total combination?
 
Physics news on Phys.org
45

10!/[(8!)(2!)]

Essentially, you have 10 choices for the first question you leave out, and 9 choices for the second question. This leads to 90 (10x9) combinations. But since it doesn't matter in which order you choose them, you've double counted (ie, you counted #1,#2 ans #2,#1 as different combinations), so divide by 2.

Njorl
 
Njorl's given you the correct answer, but to expand this is a simple cominatrics question, your calculator will probably have a function that allows you to work it out, combinations are given by:

[tex]_nC_r \equiv \left(\begin{array}{c}n\\r\end{array}\right) \equiv \frac{n!}{r!(n-r)!}[/tex]

that is n different things taken r at a time, so in this case n = 10 and r = 8.
 
Last edited:

Similar threads

Undergrad Probability to get a certain number or less by throwing two 6-faced die
  • · Replies 6 ·
Replies
6
Views
3K
High School Probability of picking one black and one white marble
  • · Replies 4 ·
Replies
4
Views
3K
Undergrad Particle clusters in a random lattice
  • · Replies 3 ·
Replies
3
Views
1K
Undergrad Choosing from identical objects of different types
  • · Replies 1 ·
Replies
1
Views
2K
High School Probability - two possible points of view?
  • · Replies 7 ·
Replies
7
Views
2K
Undergrad Please Explain (actually explain) The Monty Hall Problem
  • · Replies 131 ·
5
Replies
131
Views
13K
High School Combinatorics and Magic squares
  • · Replies 2 ·
Replies
2
Views
3K
High School Baccarat math and beating the edge
  • · Replies 3 ·
Replies
3
Views
5K
High School Can You Calculate the Odds of These Unlikely Events?
  • · Replies 8 ·
Replies
8
Views
2K
Graduate Probability puzzle
  • · Replies 21 ·
Replies
21
Views
1K